Question

Let $L$ be the line with slope $\frac{5}{12}$ that contains the point $A = (24, -1),$ and let $M$ be the line perpendicular to line $L$ that contains the point $B = (5, 6).$ The original coordinate axes are erased, and line $L$ is made the $x$ -axis and line $M$ the $y$ -axis. In the new coordinate system, point $A$ is on the positive $x$ -axis, and point $B$ is on the positive $y$ -axis. The point $P$ with coordinates $(-14, 27)$ in the original system has coordinates $(\alpha, \beta)$ in the new coordinate system. Find $\alpha + \beta.$

Answer

Answer: 31

Concepts:coordinate geometry distance formula transformation

Difficulty rating: 2390

Solution:

Line $L$ is $5x - 12y - 132 = 0$ and line $M$ is $12x + 5y - 90 = 0.$ The new $x$ -coordinate of a point is its signed distance to line $M,$ counted positive on the side containing $A,$ and the new $y$ -coordinate is its signed distance to line $L,$ positive on the side containing $B.$

Substituting $P = (-14, 27)$ into $12x + 5y - 90$ gives $-168 + 135 - 90 = -123,$ while $A$ gives $193 \gt 0;$ dividing by $\sqrt{12^2 + 5^2} = 13,$ we get $\alpha = -\frac{123}{13}.$ Substituting $P$ into $5x - 12y - 132$ gives $-70 - 324 - 132 = -526,$ and $B$ gives $-179,$ so $P$ lies on the same side of $L$ as $B$ and $\beta = \frac{526}{13}.$

Therefore $\alpha + \beta = \frac{-123 + 526}{13} = \frac{403}{13} = 31.$

2011 AIME I Problem 3

Answer: 31

Difficulty rating: 2390

Solution:

Problem 3 in Other Years